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3 years ago

X: ax squared = bx? can someone help me

Mathematics

1 answer:

SSSSS [86.1K]3 years ago

6 0

Step-by-step explanation:

ax² = bx

ax² − bx = 0

x (ax − b) = 0

x = 0 or b/a

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Brainliest and 20 pints please help

morpeh [17]

Step-by-step Explanation:

Part A: Algebraically write the equation of the best fit line in slope-intercept form. Include all of your calculations in your final answer.

Considering the data

Age (in years) Weight (in pounds)

2 32

6 47

7 51

4 40

5 43

3 38

8 60

1 23

Considering the slope intercept form

$y = mx + b$

where $m$ is the slope of the line and $b$ is the y-intercept.

Taking two points, lets say (2, 32) and (8, 60)

$\mathrm{Slope\:between\:two\:points}:\quad \mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}$

$\left(x_1,\:y_1\right)=\left(2,\:32\right),\:\left(x_2,\:y_2\right)=\left(8,\:60\right)$

$m=\frac{60-32}{8-2}$

$m=\frac{14}{3}$

$y = mx + b$

$\:\:32\:=\:\left(\frac{14}{3}\right)2\:+\:b$

$\mathrm{Switch\:sides}$

$\left(\frac{14}{3}\right)\cdot \:2+b=32$

$b=\frac{68}{3}$

Thus

$y = mx + b$

$m=\frac{14}{3}$ and $b=\frac{68}{3}$

substituting in the slop-intercept form

$y=\left(\frac{14}{3}\right)x+\left(\frac{68}{3}\right)$

Therefore, $y=\left(\frac{14}{3}\right)x+\left(\frac{68}{3}\right)$ is the equation of the best fit line in slope-intercept form.

Part B: Use the equation for the line of best fit to approximate the weight of the little girl at an age of 14 years old.

As the equation of line is

$y=\left(\frac{14}{3}\right)x+\left(\frac{68}{3}\right)$

So in order to find the weight of the little girl at an age of 14 years old, put

$x = 14$ in the above equation.

$y=\left(\frac{14}{3}\right)x+\left(\frac{68}{3}\right)$

$y=\left(\frac{14}{3}\right)\left(14\right)+\left(\frac{68}{3}\right)$ ∵ $x = 14$

$\:y=88$

Therefore, the approximate weight of the little girl at an age of 14 years old will be 88 pounds.

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kolezko [41]

Answer:

-16

Step-by-step explanation:

2*-3+5*-2=-16 Hope this helps :D

Please mark brainliest

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